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The function $f: \mathbb{R} \longrightarrow \mathbb{R}$ is defined by the rule $$f(x,y) = \begin{cases} \frac{x^5y}{x^4+y^2}, & (x,y) \neq (0,0), \\ 0, & (x,y)=(0,0). \end{cases}$$ Evaluate $f_x(0,y)$, $f_x(0,0)$, $f_y(x,0)$, and $f_y(0,0)$. (The definition of partial derivative as a limit is the recommended method.)

My question is really a conceptual question, that I am struggling to understand. At the point $(0,0)$, the function is defined to be $0$. So, finding the partial derivatives at those points, why would you not use $f(x,y)=0$?

My instincts tell me that no, you would use the other part of the function (partly because the question suggests using the limit definition): $\dfrac{x^5y}{x^4+y^2}$, $(x,y) \neq (0,0)$. But why is this?

Secondly, in general, why would you use the limit definition of a derivative, as opposed to simply taking the derivative mechanically? What is the difference?

Thanks for any help!

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When you compute the derivative of a single-variable function, it is not just the value of the function at the point that matters, but also the values of the function near but not at the point. You use both when you compute $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$Here you use the limit because "taking the derivative mechanically" requires a single formula for the function in a neighborhood of the point in question, which you don't have here. Think about how you would compute the derivative at $0$ of the function $f(x)$ given by $f(x) = x^2\sin(1/x)$ if $x\neq 0$ and $f(0)=0$. –  Arturo Magidin Jun 6 '12 at 4:46
    
@ArturoMagidin I have struggled for weeks to grasp this concept, thank you so much for explaining it clearly to me. Actually, the example you gave, is almost the same question I encountered that I struggled to answer a few weeks ago. I couldn't understand why I wouldn't just take the derivative, as opposed to using the limit definition. Now I get it! Thanks! –  JackReacher Jun 6 '12 at 5:09
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First: Your instinct is correct. Derivatives are "local information." This means that the derivative of a function at any point is computed by using information about the function's nearby values. For example, consider the limit definition of a derivative of a single-variable function: $$f'(x) = \lim_{h\to x}\frac{f(h) - f(x)}{h - x}.$$ In order to compute this, we need to understand all of the values $f$ takes at the $h$ values near $x$.

Second: Taking the derivative "mechanically" is just cranking through a set of rules that you have learned for some particularly nice functions (products, sums, quotients of differentiable functions; exponentials, trig functions, polynomials, etc). You can't take the derivative at $0$ mechanically here because, at zero, your function is not one of those particularly nice functions.

Since you can't use the differentiation rules, you're stuck with computing the derivative straight from the limit definition.

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so basically, because the function is different for the values nearby to 0, I can't just take the derivative at that point. –  JackReacher Jun 6 '12 at 5:11
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You can take the derivative, but not with the derivative rules. The problem isn't that the function has different values near $(0,0)$. Just being defined piecewise, like $f$, means that at the points where the definition changes pieces, it's not any of the functions the differentiation rules apply to, so you can't use them. –  Neal Jun 6 '12 at 5:33
    
Sure sure, thats what I meant, the function is different for values nearby to (0,0) compared to how the function is defined at (0,0), not that the values are different. Thanks again! –  JackReacher Jun 6 '12 at 5:40
    
I have another question if you can explain to me: when I use the limit definition of the partial derivative for fx(0,0) and fy(0,0), I get a 0/0 term. I know the answer to both is 0, but how would I deal with the 0/0 terms? –  JackReacher Jun 6 '12 at 10:19
    
@mathstudent You might be skipping some extra algebraic simplification. Looking at it, for $f_y$ and for $f_x$ both at $(0,0)$, I don't see $0/0$. –  Neal Jun 6 '12 at 12:38
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